Question

How do you evaluate `1/2 (sin (pi/2) + sin (pi/6))`?

Answer 1

3/4

Explanation:

Call the product `P = (1/2)p`
Apply the trig identity:
`sin a + sin b = 2sin ((a + b)/2).cos ((a - b)/2)`.
We get:
`sin ((a + b)/2) --> sin ((pi/2+ pi/6)/2) = sin (pi/3)`
`cos (a - b)/2 --> cos (pi/2 - (pi)/6)/2 = cos (pi/6)`
`p = sin pi/2 + sin pi/6 = 2sin(pi/3).cos (pi/6)`
Trig table gives -->
`sin (pi/3) = sqrt3/2`, and `cos (pi/6) = sqrt3/2`
Therefor,
p = sin (pi/2) + sin (pi/6) = 2(sqrt3/2)(sqrt3)/2 = 3/2
Finally:
P = (1/2)p = (1/2)(3/2) = 3/4